Problem: A curve in the plane is defined parametrically by the equations $x=2\sin(1+3t)$ and $y=2t^3$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{t^2}{\cos(1+3t)}$ (Choice B) B $\dfrac{3t^2}{\cos(1+3t)}$ (Choice C) C $6t^2$ (Choice D) D $36t^2\cos(1+3t)$
Solution: In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=2\sin(1+3t)$ and $y=2t^3$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(2t^3\right)}{\dfrac{d}{dt}(2\sin(1+3t))} \\\\ &=\dfrac{6t^2}{6\cos(1+3t)} \\\\ &=\dfrac{t^2}{\cos(1+3t)} \gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{t^2}{\cos(1+3t)}$.